Don't Be Tricked!

Algebra Level 1

$\small \dfrac { { (x-1) }(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10) }{ (x-1)(x-3)(x-4) } =0$

Type the number of solutions for $x$ in the above equation.

The answer is 7.

2 solutions

First Last
Dec 14, 2015

There are 10 solutions to make the numerator 0 (x = 1...10). But 1, 3, and 4 do not work as they make the denominator zero as well. 0/0 is then indeterminate, making only 7 solutions.

Wow, you're all quite intelligent. 7's just my favorite number.

Zienne Dampier - 4 years, 8 months ago

Or you can just cancel the top+bottom for the terms.

Brian Wang - 5 years, 6 months ago

It's mathematically incorrect do this, unless u're talking 'bout limits.

I can't say, for example, that x/x is always 1, 'cause if x = 0, we would have a 0/0 result, witch is indeterminate.

I can use limits and prove that lim x->0 x/x goes to 1, but the question doesn't ask for limits.

So, be careful with those things.

Victor Veras - 5 years, 5 months ago

I meant cancel top and bottom for both terms, where x is not equal to 1 3 or 4 (Because those aren't solutions anyways)

Brian Wang - 5 years, 5 months ago

Youssef Hassan F
Dec 24, 2015

by cancelling x-1 x-3 x-4 only x-2 x-5 x-6 x-7 x-8 x-9 and x-10 will remain and by making x = 2 or 5 or 6 or 7 or 8 or 9 or 10 the result will be zero due to the rule that any number × 0=0 then there are only 7 solution to this equation

Don't Be Tricked!

The answer is 7.

2 solutions

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